Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,047 questions , 2,200 unanswered
5,345 answers , 22,709 comments
1,470 users with positive rep
816 active unimported users
More ...

  Fourier-like expansion of a closed curve in 2D

+ 3 like - 0 dislike
2339 views

Fourier expansion can be used to represent any periodic function in one variable.

Closed surfaces in 3D can be built out of spherical harmonics.

Is there a similar expansion to represent a curve of any shape, like the following one?

enter image description here

This post imported from StackExchange Mathematics at 2014-06-02 11:01 (UCT), posted by SE-user Revo
asked Nov 18, 2011 in Mathematics by Revo (260 points) [ no revision ]
Are you thinking of the curve as a function y(x) ? If so, then it's not a single valued function of x, so the normal elementary Fourier transform cannot be applied. If, however, you were thinking of functions ON the curve, then they'd be periodic by construction so you could Fourier transform them.

This post imported from StackExchange Mathematics at 2014-06-02 11:01 (UCT), posted by SE-user twistor59
I think this may be a good candidate for migration to Mathematics... thoughts?

This post imported from StackExchange Mathematics at 2014-06-02 11:01 (UCT), posted by SE-user David Z
A classic example to this is "epicyles" used to calculate planets trajetories in prekeplerian times. AFAIK this method is still used in in practical astronomy (directing telescopes, space flight) because it is much easier to program.

This post imported from StackExchange Mathematics at 2014-06-02 11:01 (UCT), posted by SE-user Georg
@Revo, your curve is a function of $(x(t), y(t))$, just do two Fourier expansions for those two functions

This post imported from StackExchange Mathematics at 2014-06-02 11:01 (UCT), posted by SE-user lurscher
@DavidZaslavsky Why it should be moved to mathematics? Some shape in space can be built out of spherical harmonics which has so many applications, similarly the concepts of multipole moments and expansion which all have physics applications. If we can expand closed surface, why cannot we expand a closed line?

This post imported from StackExchange Mathematics at 2014-06-02 11:01 (UCT), posted by SE-user Revo
@twistor59 I never said the closed curve can be represented by a single valued function, but anyway can a double valued function be expanded in terms of some set of "orthogonal double valued functions"?

This post imported from StackExchange Mathematics at 2014-06-02 11:01 (UCT), posted by SE-user Revo
@Revo, sure, which are better known as just do two Fourier expansions for those two functions

This post imported from StackExchange Mathematics at 2014-06-02 11:01 (UCT), posted by SE-user lurscher
You could think of the path of the curve as being in the complex plane. Then you can do the Fourier transform of that complex function.

This post imported from StackExchange Mathematics at 2014-06-02 11:01 (UCT), posted by SE-user Greg P
@Revo (4 comments up): having physics applications is not a sufficient condition to be on topic on Physics.SE. I'm going ahead and migrating this.

This post imported from StackExchange Mathematics at 2014-06-02 11:01 (UCT), posted by SE-user David Z
demonstration: youtube.com/watch?v=QVuU2YCwHjw

This post imported from StackExchange Mathematics at 2014-06-02 11:01 (UCT), posted by SE-user Mark Eichenlaub
Since Georg brought up epicycles: have a look at this.

This post imported from StackExchange Mathematics at 2014-06-02 11:01 (UCT), posted by SE-user J. M.

1 Answer

+ 4 like - 0 dislike

As lurscher suggests in a comment, in the case of a closed curve, one could consider a periodic parametrization of the curve

$${\bf f}(\theta)~=~{\bf f}(\theta+2\pi)~\in~\mathbb{R}^2, \qquad {\bf f}(\theta)~=~(x(\theta),y(\theta)). $$

Then define Fourier coefficients in the standard way

$$ {\bf c}_n({\bf f})~:=~ \int_0^{2\pi} \frac{{\rm d}\theta}{2\pi} e^{-in\theta}~{\bf f}(\theta). $$

(The Fourier coefficients ${\bf c}_n({\bf f})$ are well-defined if the coordinate functions $x,y$ are Lebesgue integrable $x,y\in{\cal L}^1(\mathbb{R}/2\pi\mathbb{Z}).$) The Fourier series for ${\bf f}$ is vector-valued

$$\sum_{n\in\mathbb{Z}}{\bf c}_n(f) ~e^{in\theta}.$$

A similar approached works also for a closed curve in higher dimensions. In the 2 dimensional case, one may identify the plane $\mathbb{R}^2\cong \mathbb{C}$ with the complex plane, as Greg P, Mark Eichenlaub, and J.M. point out.

This post imported from StackExchange Mathematics at 2014-06-02 11:01 (UCT), posted by SE-user Qmechanic
answered Nov 18, 2011 by Qmechanic (3,120 points) [ no revision ]
We can also think of it as just a usual complex-valued Fourier transform, since complex numbers can represent two dimensions. (I now see that Greg P pointed this out in the comments to the main question.)

This post imported from StackExchange Mathematics at 2014-06-02 11:01 (UCT), posted by SE-user Mark Eichenlaub
One can also take $\mathbf f(\theta)=x(\theta)+i\,y(\theta)$; i.e., consider a complex-valued function instead of a vector-valued function.

This post imported from StackExchange Mathematics at 2014-06-02 11:01 (UCT), posted by SE-user J. M.

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsOverflo$\varnothing$
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...